Question

Use the graph of y=f(1/x) to find lim (x->infinity) f(x) and lim (x->negative infinity) f(x)
f(x)=xe^x

Answer 1

Calculus is a wonderful thing.. T_T

Answer 2

giv meh a sec ok :)

Answer 3

Thanks a lot :D

Answer 4

its a pleasure :)

Answer 5

http://www.sagemath.org/calctut/inflimits.html
mayb this will help

Answer 6

we need the graph

Answer 7

wow nm im blind...

Answer 8

There is no graph

Answer 9

ohhhh so u want a graph calculator

Answer 10

That they give us

Answer 11

No, what I mean is that I don't really understand what they are asking me to do

Answer 12

it asks you to use the graph of \(f(\frac{1}{x})=(\frac{1}{x})e^{\frac{1}{x}}\)

Answer 13

I dont understand how the graph of f(1/x) will help me find the other two limits

Answer 14

how does tha graph look like

Answer 15

do they giv u a pic of tha graph

Answer 16

They don't show us. They want us to use a graphing calculator to find it. I can do that, I just don't understand what they want us to do with it

Answer 17

I will take a picture of it

Answer 18

hold on a min I think I have a graph calculater 2 do tha job :)

Answer 19

let meh check first tho

Answer 20

This is what the problem looks like:
#49

Answer 21

This may help. I used geogebra to graph.

Answer 22

I appreciate it, but it doesn't really help much. There's only one function visible there. My main question is what they want me to do with the graph of f(1/x) to help me find the two limits

Answer 23

You start with f(x) = x*e^x

Answer 24

then you replace every copy of x with 1/x to get

f(1/x) = (1/x)*e^(1/x)

Answer 25

which is g(x) in that graph posted

Answer 26

OH MY GOD, I AM SO STUPID xD

Answer 27

I realized that just before I saw what you typed LOL

Answer 28

Ok, so I have another question

Answer 29

http://www.freemathhelp.com/factoring-calculator.php
this page has calculus calculator on it :)

Answer 30

hope it helps

Answer 31

Why does the limit of f(x) as x approaches infinity equal the limit of f(1/x) as x approaches 0 from the right side? http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/76/exercises/49/

Answer 32

giv meh a sec

Answer 33

Thanks!!

Answer 34

look at the graph of f(x) and g(x)

what is the limit as x approaches infinity for f(x)?

Answer 35

As x approaches infinity, the limit is infinity

Answer 36

The basis of one-sided limits is that when a function jumps suddenly from one value to another, it often is not possible to describe the function's behavior with a single limit

Answer 37

if that doesn't help here is a website
http://www.sagemath.org/calctut/onesided.html

Answer 38

Yeah but my question is why does the limit of the one function equal the limit of the other one
What I mean is where it says based on the example above on this site http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/76/exercises/49/

Answer 39

*based on the graph above

Answer 40

When it says the two limits of f(x) equal limits of f(1/x)

Answer 41

OH WAIT

Answer 42

as x approaches infinity, x*e^x gets extremely large so f(x) approaches infinity

as x approaches 0 from the right, (1/x)*e^(1/x) gets larger and larger, effectively approaching infinity

So that's why

\[\Large \lim_{x \to \infty}(f(x)) = \lim_{x \to 0^{+}}(g(x)) = \infty\]

Answer 43

Yes! That is exactly what I was going to say!! Thank you so much!!

Answer 44

I miss precalc and trig xD

Answer 45

you're welcome

Answer 46

I don't know this stuff sooo :/ srry bout takin u around tha block @TheWaffleMan149 :/

Answer 47

Nah, don't worry! I really appreciate the help!

Answer 48

its a pleasure :) ig :/ lol

Answer 49

do I still get a waffle 4 hard wrk lol

Answer 50

a minute agoDelete

Answer 51

Whoops

Answer 52

soooo can I have 1 pls *~*

Answer 53

Actually, i have one more question

Answer 54

One more question actually. Why does it only say limc(x) as x approaches negative infinity equal f(1/x) as x approaches 0 from the left? Doesn't f(1/x) also approach 0 when x approaches infinity?

Answer 55

*limf(x)

Answer 56

ummmm
giv meh a sec

Answer 57

Note that some functions, such as f(x) = x, do not tend to a nite value as x becomes very large, but instead
grow to innity. Other functions, such as
1
x2 , become very large even for nite values of x. So we will dene
innite limits to cover those cases:

Answer 58

*define infinite limits

Answer 59

does that help @TheWaffleMan149

Answer 60

I appreciate the effort, but not really

Answer 61

im sooooo srry idk nothing bout calculus :/
but I think ik some1 that does let meh get him 4 u

Answer 62

Don't worry about it! I appreciate you trying to help me!!!

Answer 63

r u sure u don't want meh 2 get another person 2 help u @TheWaffleMan149 :(

Answer 64

So you're asking why

\[\Large \lim_{x \to -\infty}(f(x)) = \lim_{x \to 0^{-}}(g(x))\]

is true?

Answer 65

Sorta. I know that is true, but I want to know why it doesn't also equal lim(g(x)) when x approaches infinity as well. On the graph it approaches the x axis in two places, so that is why I am confused that this only accounts for when it approaches 0 from the left

Answer 66

If that makes sense at all

Answer 67

As x approaches negative infinity, f(x) approaches 0 (you see this when you graph f(x) = x*e^x)

As x approaches 0 from the left, g(x) approaches 0. This is shown in that graph I posted above.

So that shows why

\[\Large \lim_{x \to -\infty}(f(x)) = \lim_{x \to 0^{-}}(g(x))\]

is true. They are both equal to 0.

Answer 68

And before, you saw how as x approached infinity, g(x) approached zero

so that means we have 3 different limits equal to 0

Answer 69

But why isn't that included in the answer? They only address the one approaching 0 from the left

Answer 70

It only says that limf(x) as x approaches infinity is equal to limf(1/x) as x approaches 0 from the left. It says nothing about both of those being equal to limc(1/x) as x approaches infinity

Answer 71

Sorry I keep reiterating this, I just have a hard time being able to tell who understands what I am saying when it is online

Answer 72

Is it because you only need to know that one of g(x)'s limits equal 0 to know that limf(x) as x approaches negative infinity equals 0?

Answer 73

"It only says that limf(x) as x approaches infinity is equal to limf(1/x) as x approaches 0 from the left."

that is false

\[\Large \lim_{x \to \infty}(f(x)) =\infty\]

while

\[\Large \lim_{x \to 0^{-}} \left(f\left(\frac{1}{x}\right)\right) = 0\]

So,

\[\Large \lim_{x \to \infty}(f(x)) \neq \lim_{x \to 0^{-}} \left(f\left(\frac{1}{x}\right)\right)\]

Answer 74

where does it say that? In the book? or solutions manual?

Answer 75

This website that I am getting this from has the answers to my book.

Answer 76

http://www.slader.com/textbook/9780132014083-calculus-graphical-numerical-algebraic-3rd-edition/76/exercises/49/

Answer 77

My book lists the answers, the website gives a little bit of info on how to find them

Answer 78

It's not a true or false question

Answer 79

The section of the website where it says "Based on the graph above" is what is confusing me. The second line of it

Answer 80

So you're not sure about the notation on the second line?

Answer 81

No, my confusion is that g(x) approaches 0 in two places, so why does the second line only acknowledge when it approaches it from the left side of 0. It also approaches it when x approaches infinity. Does that make any sense?

Answer 82

I see. Well on the second line when they have 0^{-}, they mean "approach 0 from the left"

So this is saying that as you approach x = 0 from the left, g(x) = f(1/x) is approaching 0.

As to why they don't mention the limit

\[\Large \lim_{x \to \infty} \left(f\left(\frac{1}{x}\right)\right) = 0\]

I'm not sure

Answer 83

Actually g(x) approaches 0 in three places: as x approaches positive infinity, negative infinity and 0 from the left.

Answer 84

Yeah. Thats my big question. Why do they only mention the 0 front the left

Answer 85

Do you think that is is because they only need one to prove that it does approach 0 and is therefore equal to lim f(x) when x approaches infinity?

Answer 86

Ok I reread the solution that you linked and I see what they want

Answer 87

They provide the graph of f(1/x) and you use this to find limits for f(x) (x approaching positive and negative infinity)

Answer 88

as x approaches 0 from the right, f(1/x) approaches infinity

When you approach 0 from the right on 1/x, this means 1/x is approaching positive infinity (x is on the right side of 0, so it's positive, 1/x is positive)

So that's where they are getting that first line under where it says "based on the graph above"

Answer 89

So that answers the first part of the question (which they show under the "result" portion)

Answer 90

I understand that one. Its the second part that I don't understand. I don't understand why they ignored the other two places where the limit is 0

Answer 91

I understand why they have what they do. My only confusion is why they left those out

Answer 92

when x approaches 0 from the left, f(1/x) is approaching 0

x is now negative, so 1/x is growing to negative infinity

therefore saying "x --> 0^{-}" is the same as "1/x ---> -infinity"

so that's how they got that second line

Answer 93

they are true facts, but they aren't needed to answer the original question

"Use the graph of y=f(1/x) to find lim (x->infinity) f(x) and lim (x->negative infinity) f(x) f(x)=xe^x"

Answer 94

we only care about what they are originally asking, which are the limits for f(x)

x is approaching positive infinity
x is approaching negative infinity

Answer 95

So like I said earlier, it is just because they only need one of the 3 to prove that it has a limit at 0?

Answer 96

I'm not sure how you can use the others, but yes, that is the key needed to prove the second line

Answer 97

Ok. Thank you so much for the help!! I really, really appreciate it :D